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butchic · Oct 20, 2009, 07:08 AM

Find the perimeter of an isosceles triangle if the vertex angle measures 50 degrees and the length of the base is 28 cm. help.tnx

s_cianci · Oct 20, 2009, 07:14 AM

Draw an altitude from the vertex angle to the base. Then use the cosine ratio to calculate the length of one of the legs (the other leg will then be the same.) Then add the base and the 2 legs together to get the perimeter.

butchic · Oct 20, 2009, 07:31 AM

Can you please show me how to do it. I'm so dumb on this.. tnx

Nhatkiem · Oct 20, 2009, 09:33 AM

Originally Posted by butchic

can you please show me how to do it. im so dumb on this.. tnx

Think about what an isosceles triangle has to have. 2 sides of equal length, and 2 angles that are equal.

The sum of those angles have to be 180 deg. However, 1 angle is already given to be 50 deg, and if 2 angles are the same, then we have that

$2\theta=180-50$

Now we know all the angles, and the length of side "c" (the side opposite to the 50 deg angle).

using the law of cosines (You can also use the law of sines, but I stick with cosines out of habit unless I can't use it, then I will switch to law of sines), we can figure out the length (which I will call a, and b) has to be. The law of cosines states

$c^2=a^2+b^2-2abcos(\theta_c)$ (I used $\theta_c$ here to represent the angle opposite of side "c".)

Remember that 2 sides have to be equal, so what can we generalize about the sides a and b, if c is known to be the base opposite of $\theta_c$ ?